The arithmetic sequence 20, 16, 12, 8, … ends with the term −120. How many terms n does it have?
Sequences & series worksheet with answers
1.Easy
a₁ = 20d = −4aₙ = −120
n =
2.Easy
Find the sum S6 of the first 6 terms of the geometric sequence 1, −2, 4, …
a₁ = 1r = −2n = 6
S6 =
3.Medium
Exactly one of the four series diverges. Click it.
A: Σ (5/3)ᵏB: Σ 5/k²C: Σ (−1)ᵏ/kD: Σ 3ᵏ/k!
- 1A: Σ (5/3)ᵏ
- 2B: Σ 5/k²
- 3C: Σ (−1)ᵏ/k
- 4D: Σ 3ᵏ/k!
4.Medium
Apply the root test to Σk=1∞ ((3k + 2)/(2k + 5))ᵏ. Find L = lim |ak|1/k.
ak = ((3k + 2)/(2k + 5))ᵏ
L =
5.Hard
f(x) = sin x is approximated by its Maclaurin polynomial of degree 2 at x = 0.2. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₂|.
f(x) = sin xn = 2x = 0.2M = 1
|R₂| =
6.Hard
Use the Maclaurin polynomial of degree 4 of f(x) = cos(3x) to approximate f(0.5). Give P₄(0.5).
f(x) = cos(3x)n = 4x = 0.5
P₄(0.5) =
Answer key
- 1. n = 36
- 2. S6 = −21
- 3. A: Σ (5/3)ᵏ
- 4. L = 1.5
- 5. |R₂| = 0.001333
- 6. P₄(0.5) = 0.08594
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